Question

Two balls (A and B) of different size have charge $-4mC$ and $-1mC$ respectively. When placed at certain gap they apply a force $20N$ on each other. Now bulb is made to touch each other. The new charge on ball A is $1mC$. Balls are again placed at same gap. Get the new force between them

• A. $10N$
• B. $5N$
• C. $20N$
• D. None

Answer 1

The correct option is D None

we know that

$\begin{array}{c}F=\frac{k{q}_1{q}_2}{r^2}\\ 20=\frac{9\times {10}^9\times \left(-4\right)\times {10}^{-3}\times \left(-1\right)\times {10}^{-3}}{x^2}\\ x^2=1800\\ x=30\sqrt{2}m\end{array}$

Now after the balls are touched, the charge on ball A is -1mC.

Applying the law of conservation of charge (i.e. the total charge must remain same throughout),

The charge on ball B is -5-1=-6mC

New force will be:

$\begin{array}{c}F=\frac{9\times {10}^9\times \left(-6\right)\times {10}^{-3}\times \left(+1\right)\times {10}^{-3}}{1800}\\ F=-30N\end{array}$

(-ve sign indicating that the force is attractive)